SOLUTION: What is the equation of a parabola with vertex on the line y=x, axis parallel to Ox, and passing through (6,-2) and (3,4)?

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Question 1045379: What is the equation of a parabola with vertex on the line y=x, axis parallel to Ox, and passing through (6,-2) and (3,4)?
Found 2 solutions by KMST, advanced_Learner:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Let the vertex be V%28k%2Ck%29 .
The equation of a parabola with that vertex and axis parallel to the x-axis is
x-k=a%28x-k%29%5E2 .
Substituting the coordinates of the given points, we have
system%28x=6%2Cy=-2%29--->6-k=a%28-2-k%29%5E2
system%28x=3%2Cy=4%29--->3-k=a%284-k%29%5E2
Those two equations form a system of equation we have to solve to find a and k .
Subtracting one equation from the other, we have
6-k-%283-k%29=a%28-2-k%29%5E2-a%284-k%29%5E2
6-k-3%2Bk%29=a%28%28-2-k%29%5E2-%284-k%29%5E2%29
3=a%28-2-k%2B%284-k%29%29%28-2-k-%284-k%29%29
3=a%28-2-k%2B4-k%29%28-2-k-4%2Bk%29
3=a%28-2k%2B2%29%28-6%29
3=a%28-2%29%28k-1%29%28-6%29
3=12a%28k-1%29
3%2F12%28k-1%29=a
1%2F4%28k-1%29=a
Substituting the expression 1%2F4%28k-1%29 for a in 3-k=a%284-k%29%5E2 we get
3-k=%284-k%29%5E2%2F4%28k-1%29-->4%28k-1%29%283-k%29=%284-k%29%5E2-->
4%28-k%5E2%2B4k-3%29=%284-k%29%5E2-->-4k%5E2%2B16k-12=16-8k%2Bk%5E2
-->0=16-16k%2Bk%5E2%2B4k%5E2-16k%2B12-->0=5k%5E2-24k%2B28
The solutions to that equation are k=2 and k=24%2F5 .
Only one of those solutions should work, because
through any 3 points, (such as (6,-2) , (3,4), and V%282%2C2%29 )
passes only one parabola with axis parallel to the x-axis.
Substituting 2 for k in the original equation
3-k=a%284-k%29%5E2 , we get
3-2=a%284-2%29%5E2--->1=4a--->a=1%2F4
The values system%28k=2%2Ca=1%2F4%29 also satisfy equation 6-k=a%28-2-k%29%5E2 .
So, highlight%28x-2=%28y-2%29%5E2%2F4%29 is the equation of a parabola with vertex on the line y=x , axis parallel to the x-axis, and passing through (6,-2) and (3,4).
The equation can be written in other, equivalent forms.
highlight%28x-2=%28y-2%29%5E2%2F4%29<-->highlight%28x=%28y-2%29%5E2%2F4%2B2%29<-->highlight%28x=%281%2F4%29y%5E2-x%2B3%29

NOTE: The solution k=14%2F5 does not yield the same a value when substituted in the two original equations.

Answer by advanced_Learner(501) About Me  (Show Source):